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Question: How many grams of \[C{{O}_{2}}\] are in \[520mL\] of carbon dioxide gas at \(0\) degrees Celsius and...

How many grams of CO2C{{O}_{2}} are in 520mL520mL of carbon dioxide gas at 00 degrees Celsius and a pressure of 750.3750.3 torr?

Explanation

Solution

We know that the ideal gas law states certain assumptions about gases and theory which are unnecessarily false. Therefore this results that ideal gas law has some sort of limitation within it. For example; ideal gas law creates an assumption that the gas particles have neither volume nor are attracted to one another.

Complete step-by-step answer:
Firstly we have to put mass of CO2C{{O}_{2}} under the given conditions 1.01g1.01g
After that we have to determine mol CO2mol\text{ }C{{O}_{2}} by using Ideal gas laws and multiplying mol CO2mol\text{ }C{{O}_{2}}by its molar mass 44.0098gmol44.0098\dfrac{g}{mol} . Whereas the ideal gas law equation is given by: PV=nRTPV=nRT.

Here we have PP is pressure, VV is volume, nn is moles, RR is the gas constant and TT is temperature in Kelvins and the gas constant includes volume of unit in liters and the volume milliliters would be converted into liters.
Given :
P = 750.3 Torr\Rightarrow P\text{ }=\text{ }750.3\text{ }Torr
V = 520mL × 1L1000mL  = 0.520 L\Rightarrow V\text{ }=\text{ }520mL\text{ }\times \text{ }\dfrac{1L}{1000mL}~~=\text{ }0.520\text{ }L
R =  62.364 L TorrKmol \Rightarrow R\text{ }=~~62.364\text{ }L\text{ }\dfrac{Torr}{Kmol}~
T = 0C + 273.15 = 273 K\Rightarrow T~=~0{}^\circ C~+~273.15~=~273\text{ }K
Now that we know the formula or as we say ideal gas equation; PV=nRTPV=nRT
Thus from here we can determine the formula for nn we get formula for n=RTPVn=\dfrac{R\cdot T}{P\cdot V} and now by substituting the values we get:
n=750.3Torr×0.520L62.364LTorrKmol×273K=0.02298molCO2\Rightarrow n=\dfrac{750.3Torr\times 0.520L}{62.364L\dfrac{Torr}{Kmol}\times 273K}=0.02298mol\cdot C{{O}_{2}}
Where mass of CO2C{{O}_{2}} can be given by, multiplying the value of nn with 44gCO21molCO2\dfrac{44gC{{O}_{2}}}{1molC{{O}_{2}}} we get;
0.02298molCO2×44gCO21molCO2\Rightarrow 0.02298mol\cdot C{{O}_{2}}\times \dfrac{44gC{{O}_{2}}}{1molC{{O}_{2}}}
1.01gCO2\Rightarrow 1.01g\cdot C{{O}_{2}}
Therefore, 1.011.01 grams of CO2C{{O}_{2}} are in 520mL520mL of carbon dioxide gas at 00 degrees Celsius and a pressure of 750.3750.3 torr.

Note: Note that since this particle of ideal gases have neither volume therefore gas should be able to get condensed to the volume of zero. Whereas real gaseous particle that occupy space. A gaseous state will be more and more condensed in order to form liquidity and has volume. The gaseous law have no longer application which is because of substance is no longer in a gaseous state.